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我正在尝试将我的应用程序的 web.xml 迁移到基于 Java 的配置。我们使用的是 spring 4.1、Java 7、Servlet 3.1、Tomcat 8 和 Eclipse Luna。Web 服务框架是 Jersey 2.14。

我主要使用以下指南：http ://www.robinhowlett.com/blog/2013/02/13/spring-app-migration-from-xml-to-java-based-config/

我创建了遵循 web.xml 配置的 WebApplicationInitializer，删除了 web.xml，将 Maven 配置为不查找 web.xml，并成功执行了 mvn clean install。

当我尝试启动 tomcat 时，出现以下错误：

'Publishing to Tomcat v8.o Server at localhost...' has encountered a problem. Resource '/sb-server/target/m2e-wtp/web-resources/WEB-INF/web.xml' does not exist.

我试图清理 tomcat 目录，但它没有帮助，看起来我错过了一些东西，因为 AFAIK Tomcat 8 应该是基于 Java 的配置友好的。

我错过了迁移的一步吗？

前 web.xml （按预期工作）：

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0"> 

    <context-param>
        <param-name>contextClass</param-name>
        <param-value>
    org.springframework.web.context.support.AnnotationConfigWebApplicationContext
        </param-value>
    </context-param>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
            com.sb.configuration.ServerConfiguration
        </param-value>
    </context-param>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <servlet>
        <servlet-name>Jersey REST Service</servlet-name>
        <servlet-class>
            org.glassfish.jersey.servlet.ServletContainer
        </servlet-class>
        <init-param>
            <param-name>javax.ws.rs.Application</param-name>
            <param-value>com.sb.configuration.RestJaxRsApplication</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.server.tracing.type</param-name>
            <param-value>ALL</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>Jersey REST Service</servlet-name>
        <url-pattern>/api/*</url-pattern>
    </servlet-mapping>

    <!-- Spring Security -->
    <filter>
        <filter-name>springSecurityFilterChain</filter-name>
        <filter-class>org.springframework.web.filter.DelegatingFilterProxy
        </filter-class>
    </filter>
    <filter-mapping>
        <filter-name>springSecurityFilterChain</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>

</web-app>

WebApplicationInitializer 实现：

@Order(Ordered.HIGHEST_PRECEDENCE)
public class WebAppInitializer implements WebApplicationInitializer {

    @Override
    public void onStartup(final ServletContext container) throws ServletException {

        // Set up application context
        final AnnotationConfigWebApplicationContext appContext = new AnnotationConfigWebApplicationContext();
        appContext.register(ServerConfiguration.class);
        container.addListener(new ContextLoaderListener(appContext));

        // Register listeners
        container.addListener(ContextLoaderListener.class);

        // Jersey Servlet configuration
        final ServletRegistration.Dynamic dispatcher =
                container.addServlet("Jersey REST Service", ServletContainer.class);
        dispatcher.setInitParameter("javax.ws.rs.Application", "com.sb.configuration.RestJaxRsApplication");

        dispatcher.setInitParameter("jersey.config.server.tracing.type", "ALL");
        dispatcher.setLoadOnStartup(1);
        dispatcher.addMapping("/api/*");

        // Filters
        final Dynamic filterRegistration = container.addFilter("springSecurityFilterChain", DelegatingFilterProxy.class);
        filterRegistration.addMappingForUrlPatterns(null, false, "/*");

    }
}
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1 回答 1

0

有一种更简单的方法可以做到这一点。

在您的 pom 文件中添加对 Servlet 3.1 的依赖项

<dependency>
    <groupId>javax.servlet</groupId>
    <artifactId>javax.servlet-api</artifactId>
    <version>3.1.0</version>
</dependency>

首先创建一个 Servlet 初始化程序。下面的例子：

public class WebAppInit implements WebApplicationInitializer {

    @Override
    public void onStartup(ServletContext servletContext) throws ServletException {
        AnnotationConfigWebApplicationContext context = new AnnotationConfigWebApplicationContext();
        context.scan(AppConfiguration.class.getPackage().getName());

        servletContext.addListener(new ContextLoaderListener(context));
        ServletRegistration.Dynamic appServlet = servletContext.addServlet("appServlet", new DispatcherServlet(context));

        appServlet.setLoadOnStartup(1);
        Set<String> mappings = appServlet.addMapping("/");

        if (!mappings.isEmpty()) {
            throw new IllegalStateException("Conflicting mappings found! Terminating. " + mappings);
        }
    }
}

您的应用程序配置还需要一些设置。下面的例子：

@Configuration
@EnableWebMvc
@ComponentScan("com.your.package.root")
public class AppConfiguration extends WebMvcConfigurerAdapter {

    @Bean
    public ViewResolver viewResolver() {
        InternalResourceViewResolver viewResolver = new InternalResourceViewResolver();
        viewResolver.setOrder(0);
        return viewResolver;
    }

    /**
     * Spring and WEB settings from WebConfAdapt.
     */

    @Override
    public void configureDefaultServletHandling(DefaultServletHandlerConfigurer configurer) {
        configurer.enable();
    }
}

除此之外，不需要更多配置。我成功地用这个启动了我的Tomcat。（但是，这是在 Tomcat 7 上）

于 2015-03-04T08:38:05.093 回答